Each amplification stage of a multistage amplifier introduces noise and phase shift into the signal it amplifies. It is, therefore, desirable to use an amplifier having few amplification stages with high gain and low noise. Currently available single-stage amplifiers implemented with short-channel field effect transistors (FETs) provide relatively low amounts of gain or exhibit poor noise performance.
FIG. 1 shows a prior art low-gain amplifier circuit 10 having a single amplification stage comprised of FETs Q.sub.1 and Q.sub.2 connected in cascode arrangement. An input signal, V.sub.i, which is developed from the current delivered by an input current source, I.sub.i, is applied to the gate terminal of Q.sub.1. A FET transistor Q.sub.3 functions as a current source for biasing diodes D.sub.3 and D.sub.4 that ensure a sufficient drain-to-source voltage across Q.sub.1.
A FET Q.sub.4 and diodes D.sub.5, D.sub.6, D.sub.7, D.sub.8, and D.sub.9 form a source follower and DC voltage level shifter. A FET Q.sub.5 functions as a current source for Q.sub.4, and diodes D.sub.1 and D.sub.2 drop a DC voltage, ensuring sufficient drain-to-source voltage for Q.sub.5. A feedback resistor, R.sub.f, connected between the drain terminal of Q.sub.5 and the gate terminal of Q.sub.1 may be used to form a transimpedance amplifier whose bandwidth is dependent on the voltage gain of the amplifier. The bandwidth of such a transimpedance amplifier is approximately equal to A.sub.v /(2.pi.R.sub.f C), where A.sub.v is the voltage gain of the amplifier and C is the total input capacitance. Each of the transistors Q.sub.1, Q.sub.2, and Q.sub.5 is a depletion-mode gallium arsenide (GaAs) metal semiconductor field effect transistor (MESFET).
A resistor R.sub.L, which is represented in FIG. 1 as a variable resistor, is connected between V.sub.dd and the drain terminal of Q.sub.2 to provide a passive load resistance for amplifier 10. A resistive load provides amplifier 10 with lower noise characteristics than those achievable with an active load consisting of a transistor connected as a current source. An active transistor load would typically have channel noise comparable to that of Q.sub.1 and, therefore, degrade the noise performance by approximately 3 dB. On the other hand, a resistive load has thermal noise that is typically three to ten times smaller than the channel noise of a transistor load. The voltage gain, A.sub.v, measured from the gate terminal of Q.sub.1 to the drain terminal of Q.sub.2, is approximately equal to the transconductance g.sub.m of Q.sub.1 multiplied by R.sub.L, as expressed in equation (1) below: EQU A.sub.v =-g.sub.m R.sub.L. (1)
It is known that the transconductance, g.sub.m, is proportional to the square root of the current I.sub.d that enters the drain of Q.sub.1, which is expressed in equation (2), below: EQU g.sub.m .alpha.(I.sub.d).sup.1/2. (2)
Equation (1) indicates that A.sub.v increases as the value of R.sub.L increases. An increase in the value of R.sub.L causes a corresponding decrease in the value of I.sub.d, which flows through R.sub.L. Equation (2) indicates, however, that a decrease in the value of I.sub.d decreases the value of g.sub.m, thereby counteracting the increase in A.sub.v caused by an increase in the value of R.sub.L. Therefore, according to equations (1) and (2), A.sub.v does not increase linearly with an increase in R.sub.L. As a consequence, the maximum value of A.sub.V is lower than desired. Moreover, as R.sub.L increases in value, the output resistance present at the drain of Q.sub.2 lowers the effective load resistance present at the drain terminal of Q.sub.2, thereby further limiting the overall gain, A.sub.v.
A practical implementation of circuit 10 that provides A.sub.v =-13 would include values of R.sub.L =500 ohms, V.sub.ss =-5 volts, V.sub.dd =+5 volts, and I.sub.d =10 milliamperes. The output impedance, R.sub.O, of common gate stage Q.sub.2 would be approximately 3.25 kilohms. Transistor Q.sub.1 would have g.sub.m =30 millimhos, gate width=500 microns, f.sub.T =15 GHz, and pinchoff voltage V.sub.p =-0.7 volts.